What are the absolute extrema of # f(x)= |sin(x) + ln(x)|# on the interval (0 ,9]?

Question

Cameron Alexander · Accepted Answer

No maximum. Minimum is #0#.

No maximum As #xrarr0#, #sinxrarr0# and #lnxrarr-oo#, so #lim_(xrarr0) abs(sinx+lnx) = oo# So there is no maximum. No minimum Let #g(x) = sinx+lnx# and note that #g# is continuous on #[a,b]# for any positive #a# and #b#. #g(1) = sin1 > 0# #" "# and #" "# #g(e^-2) = sin(e^-2) -2 < 0#. #g# is continuous on #[e^-2,1]# which is a subset of #(0,9]#. By the intermediate value theorem, #g# has a zero in #[e^-2,1]# which is a subset of #(0,9]#. The same number is a zero for #f(x) = abs(sinx+lnx)# (which must be non-negative fo all #x# in the domain.)

Adam Adkins · Answer

undefined To find the absolute extrema of $ f(x) = |\sin(x) + \ln(x)| $ on the interval $(0, 9]$, we need to examine the critical points and the endpoints of the interval.

1. Critical points occur where the derivative is zero or undefined.
2. Endpoints are where the function is evaluated at the boundaries of the interval.

First, find the critical points by finding where the derivative equals zero or is undefined. Then, evaluate the function at those critical points and at the endpoints of the interval. Finally, compare these values to determine the absolute extrema.

The derivative of $ f(x) $ is:
$$ f'(x) = \frac{\sin(x)}{|sin(x) + \ln(x)|} \cdot \cos(x) + \frac{1}{x \cdot |sin(x) + \ln(x)|} $$

There is a critical point where $ f'(x) = 0 $ or is undefined. However, the function $ |sin(x) + \ln(x)| $ never equals zero for any $ x $ in the given interval.

Thus, we only need to evaluate $ f(x) $ at the endpoints of the interval:

1. At $ x = 0 $, $ f(x) $ is undefined.
2. At $ x = 9 $, $ f(9) = |\sin(9) + \ln(9)| $.

Therefore, the absolute extrema of $ f(x) $ on the interval $(0, 9]$ is $ f(9) $.

Axel Alvarez · Answer

undefined The absolute extrema of $f(x) = |\sin(x) + \ln(x)|$ on the interval $(0, 9]$ are as follows:

The absolute maximum occurs at $x = \pi$ with a value of $f(\pi) = |\sin(\pi) + \ln(\pi)| = |\ln(\pi)|$.

The absolute minimum occurs at $x = e$ with a value of $f(e) = |\sin(e) + \ln(e)| = |\sin(e) + 1|$.